Exploration 2: Products
November 9, 2023.Let’s study subgroups. Let’s say and are subgroups of . How much of is captured by these two subgroups? In other words, how much of can you reconstruct from and alone?
One obvious answer is to combine the two subgroups. We can take the group product , often just called the product, as all elements for all . It is not necessarily a group. Of course this only works because and are subgroups of the same group, which gives us the definition of a product between elements of and . If and were arbitrary groups, we would have to come up with a new definition of product between them. More on that later.
contains , and it also contains . To see this, remember that the identity exists in both and , so includes the elements for all and for all . Thinking about group order, this means is at least as large as and . And is at most as large as , since elements of come from and therefore are contained in . So:
Verify that when , , and when , . The question is, when is ? In other words, when do and capture all the information about ? For this to be true, every element of must be expressible in the form for some . Since there are choices for and choices for , there are choices for , which must be at least in order to represent all of . So, firstly, in order for to be true, we must have at the very least:
Secondly, we would like to know how many of these represent unique elements of . To do this, we need to explore when two are equal, and we can take their equivalence classes as the unique elements of .
- Consider all pairs . There are of them, each corresponding to a product . But not all products are necessarily distinct: holds for every element .
- This means the equivalence class of pairs whose product is equal to consists of pairs. Dividing the number of pairs by the size of each equivalence class gives the number of distinct products .
Corollary: If and have trivial intersection, .
So in order to have , we can consider when have trivial intersection.
By the previous theorem , when we get . The assumption then gives . But we always have and therefore , thus . Since all the distinct elements in can only come from , .
Conversely, if , then and every element of is expressible by a unique pair for and , implying that using the same argument in the above theorem. This means , which together with implies , which is enough to show as well.
Proving that requires knowing and , which isn’t often the case, so let’s go in a different direction.
Whenever is trivial, we can make an argument about commutativity between elements and . In particular, if we can prove that their commutator is in both and , then we know and therefore all commute with all . When is this true?
Note that implies , which exactly means that is invariant under conjugation by elements of . By a similar argument, implies which exactly states that is invariant under conjugation by elements of . We can make both statements true if we assert that and are invariant under conjugation by elements in — in other words, and are normal subgroups of .
Then the logic goes like this. If and are normal subgroups of , then is in both and , implying that . If is trivial, then meaning that every element of commutes with every element of . Then:
We can prove two facts about the commutator of arbitrary and . - Since is normal, , so . - Since is normal, , so . Therefore the commutator is in both and . Since their intersection is trivial, the commutator must be the identity every time, implying that all elements of commute with all elements of .
In this section, we try a different approach to constructing a group from given subgroups.
If and are both normal with trivial intersection, their group product is sometimes called the internal direct product.
In particular, if , then every single is a unique element of , and together with we get . That means every pair for and can be identified with a unique element of . In fact, every element of the group can be written as a pair . In general, if two groups are renamings of each other, then they are isomorphic, and the renaming is an isomorphism. Here, we say that and are isomorphic (denoted ), and that the bijective renaming is an isomorphism.
This method of taking all pairs for and works for arbitary groups , and is known as the direct product of groups, and is denoted . It is always a group because product and inverse can be defined componentwise, for example .Sometimes the direct product is also called the direct sum, denoted , but that is specifically for abelian groups whose group operation is denoted as addition ().
Note that these three conditions are exactly the behavior of in :
- implies , thus their intersection is trivial
- implies and therefore
- is just .
That proves the forward direction. To show the reverse direction, we can construct this isomorphism directly with the given conditions.
A common misconception is that the direct product is kind of like the inverse of a group quotient. However, in general.
- Take the example .
- Let . since is cyclic, therefore abelian, therefore normal.
- Then sends to , and the result is isomorphic to .
- But , since has an order element while doesn’t.
Here are some other interesting theorems about direct product.
There’s a simple bijection between each and (swap the elements), therefore the two groups are isomorphic.
Proving this for is enough to prove the theorem by induction. Since every element of chooses an element and an element , there are possible choices for and possible choices for , so there are elements in , as required.
- Given and , we want to show .
- is generated by if every one of its elements can be written as for some .
- This amounts to solving the system for .
- Chinese Remainder Theorem: In the above equations, has only one solution mod .
- This is a number-theoretic proof.
- Bezout’s identity: if , then there exist integers such that .
- Then has a solution:
- To prove the solution unique mod , consider another solution where and .
- But then and , therefore must be a multiple of both and .
- Since , must be a multiple of , therefore and .
- Then there is exactly one solution for for every pair . So we can write a bijection between each and the corresponding , which means the two groups are isomorphic.
The direct product discussed above is known as the external direct product . If both groups are actually subgroups of some encompassing group , then we can define the internal direct product , whose elements are all products for . Unlike the external direct product, the internal direct product is restricted to working with elements of the encompassing group. Let’s explore the case when :
- ()
- implies that every can be written as for some .
- To prove that this factoring is unique given , let . Then: shows that any two such factorings are the same.
- ()
- Given that is a unique factoring for all , let be an arbitrary element in .
- Then has two factorings , which must be the same due to the unique factoring in . Then , implying the only element in the intersection is .
By the previous theorem, and implies any element can be factored uniquely into where . This unique factorization essentially maps each to a distinct pair . When , this describes an isomorphism .
In this section, we study the semidirect product.
For arbitrary subgroups of , recall that is not necessarily a subgroup.
The typical counterexample is with no relations between and (each string consisting of copies of represents a unique element.) It cannot be closed under inverses since , and is distinct from any .
- Both subgroups contain , so their product contains .
- Both subgroups are closed under product. To see this, take an arbitrary element . Since either or is normal, is equal to some (if is normal) or some (if is normal). Either way, we obtain some which is in , so is closed under product.
- Both subgroups are closed under inverses. To see this, take an arbitrary element , whose inverse is . Since either or is normal, is equal to some (if is normal) or some (if is normal). Either way, we obtain some which is in , so is closed under inverse.
- Since the product contains identity, and is closed under product and inverses, it is also a subgroup of .
Essentially, the reason is in if say is normal is because the normality of gives us, for each , a bijection defined by . This is the same as , which lets us turn every into , making closed under product as required.
Note that the key part that makes a subgroup is not the normality of , but the bijection it defines for every . This bijection lets us turn every instance of into for some element . Above, the bijection was simply conjugation by , but is a subgroup as long as we can define . To see this clearly, suppose you have again, and we want to prove it is in . Then since every instance of is equal to for some , we get and we are done since and .
The proof of this is similar to the previous one.
- Both subgroups contain , so their product contains .
- Both subgroups are closed under product. Then take an arbitrary element . This element is equivalent to
- Since either or is normal, is equal to some (if is normal) or some (if is normal). Either way, we obtain some which is in , so is closed under product.
- Both subgroups are closed under inverses. Then take an arbitrary element , whose inverse is . Since either or is normal, is equal to some (if is normal) or some (if is normal). Either way, we obtain some which is in , so is closed under inverse.
- Since the product contains identity, and is closed under product and inverses, it is also a subgroup of .
Just like with the direct product, we can generalize this to that are not subgroups of the same group. Define the semidirect product as the same thing as the direct product , except the product is not but instead . The idea for the notation is that it contains a small that indicates that acts ‘normally’ in the sense that exists for each . Indeed:
This means that
This lets us define a classification theorem for semidirect products. IF for subgroups where , and is trivial then is isomorphic to where is conjugation.
TODO wreath product?
zappa-step product?
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